226 ANALYTICAL MECHANICS Thus we have from either method ma ma It is evident from the last two equations that both the linear and the-angular accelerations are constant. Therefore the equations of motion are ma v = ma2 DISCUSSION. When / <^C ma, then v = 0, and the motion of the suspended body is about the same as that of a freely falling body. When I ^>ma then v = 0. Therefore the velocity of the falling body changes very slowly. 3. A uniform rectangular trapdoor, which is held in a vertical position, is allowed to fall. Supposing the hinges to be smooth and horizontal, find the expression for the angular velocity at any instant of the motion. ma*' ma 2 I + ma*^ *-27-& FIG. 113. TORQUE METHOD.The torque on the door is due to the action of its weight and the reaction of the hinges. Therefore n & G = mg.~ Putting this value of G in the torque equation we get ENERGY METHOD. In turning through an angle 6 the door acquires a kinetic energy of f Io>2 and loses from its potential energy an amount equal to mgh. Therefore the energy equation gives - /co2 = mgh = mg ~ (1 - cos 6). A L Differentiating with respect to the tune or mga sin 8.